How do you write the rectangular coordinates for the point: #(6, 3/2 pi)#? Precalculus Polar Coordinates Converting Coordinates from Polar to Rectangular 1 Answer 1s2s2p Mar 20, 2018 #(-3,3sqrt(3))# Explanation: For given coordinates #(r,theta)#, #(r,theta)to(x,y)=>(rcostheta,rsintheta)# #r=6# #theta=(2pi)/3# #(6,(2pi)/3)to(x,y)=>(6cos((2pi)/3),6sin((2pi)/3))-=(-3,3sqrt(3))# Answer link Related questions What is the formula for converting polar coordinates to rectangular coordinates? How do I convert polar coordinates #(5, 30^circ)# to rectangular coordinates? How do I convert polar coordinates #(3.6, 56.31)# to rectangular coordinates? How do I convert polar coordinates #(10, -pi/4)# to rectangular coordinates? How do I convert polar coordinates #(4,-pi/3)# to rectangular coordinates? How do I convert polar coordinates #(6, 60^circ)# to rectangular coordinates? How do I convert polar coordinates #(-4, 230^circ)# to rectangular coordinates? What is the Cartesian equivalent of polar coordinates #(sqrt97, 66^circ)#? What is the Cartesian equivalent of polar coordinates #(2, pi/6)#? What is the Cartesian equivalent of polar coordinates #(7, pi)#? See all questions in Converting Coordinates from Polar to Rectangular Impact of this question 2811 views around the world You can reuse this answer Creative Commons License